極限題目...............................

As follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jul07/Crazylimit1.jpg

lim x to infinity {cos[(2004+x)^0.5]-cos(x)^0.5}

For a very large x,
(2004+x)^0.5=x^0.5 (1+2004/x)^0.5
=x^0.5(1+1002/x) First order appox.

1002/x is very small


cos[(x^0.5)(1+1002/x)]=
=cos(x^0.5) + d cos (^0.5) / dx 1002/x [ f(x+dx)=f(x)+f'(x)dx ]
=cos(x^0.5) - sin(x^0.5)/[2(x^0.5)] 1002/x

lim x to infinity {cos[(2004+x)^0.5]-cos(x)^0.5}
=lim x to infinity {cos(x^0.5) - sin(x^0.5)/[2(x^0.5)] 1002/x-cos(x)^0.5}
=lim x to infinity {- 501sin(x^0.5)/(x^1.5) }

since -1

2007-07-20 16:10:38 補充：
since -1

2007-07-20 16:11:05 補充：
since -1=< sin x= <1for x trends to infinity, 1/(x^1.5) becomes zeroso lim x to infinity {- 501sin(x^0.5)/(x^1.5) } = 0We have lim x to infinity {cos[(2004+x)^0.5]-cos(x)^0.5} = 0